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Homework Help: Prove using the Triangle Inequality

  1. Sep 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that:

    (|x+y|)/(1+|x+y|) ≤ ((|x|)/(1+|x|)) + ((|y|)/(1+|y|))

    2. Relevant equations

    You are given the triangle inequality:

    |x+y| ≤ |x| + |y|

    3. The attempt at a solution

    (This is done from the result, as I haven't been able to find the starting point)

    (|x+y|)/(1+|x+y|) ≤ (|x|(1+|y|)+|y|(1+|x|))/((1+|x|)(1+|y|))

    (|x+y|)/(1+|x+y|) ≤ (|x|+2|x||y|+|y|)/(1+|x|+|y|+|x||y|)

    This doesn't seem to go anywhere. I also tried flipping the whole thing to get:


    but this doesn't seem to lead anywhere either...

    I'm not sure how to go about this problem.
  2. jcsd
  3. Sep 23, 2011 #2
    As a first step, try starting by applying the triangle inequality to the numerator of the left-hand side. Next, what can you say about [itex] 1 + |x+y|[/itex] and [itex] 1+|x|[/itex]? What does this imply about [itex] \frac{1}{1+|x+y|}[/itex] and [itex] \frac{1}{1+|x|}[/itex]?
  4. Sep 24, 2011 #3
    so, since 1+|x+y| [itex]\geq[/itex] 1+|x|,

    [itex]\frac{1}{1+|x+y|}[/itex] [itex]\leq[/itex] [itex]\frac{1}{1+|x|}[/itex]


    [itex]\frac{|x+y|}{1+|x+y|}[/itex] [itex]\leq[/itex] [itex]\frac{|x|}{1+|x+y|}[/itex] + [itex]\frac{|y|}{1+|x+y|}[/itex]


    [itex]\frac{|x|}{1+|x+y|}[/itex] + [itex]\frac{|y|}{1+|x+y|}[/itex] [itex]\leq[/itex] [itex]\frac{|x|}{1+|x|}[/itex] + [itex]\frac{|y|}{1+|x+y|}[/itex]

    Can a similar argument now simply be extended for 1+|y|?
  5. Sep 24, 2011 #4
    You're on the right track. Prove the inequality is true for the denominator, then you can easily prove it is true for the numerator.
  6. Sep 24, 2011 #5
    Actually, this isn't true: take x=1 and y=-1 for a counterexample.

    I just noticed a trick that makes this much easier:

    \frac{a}{1+a} = \frac{1+a-1}{1+a} = \frac{1+a}{1+a} - \frac{1}{1+a} = 1 - \frac{1}{1+a}\; .

    Try using this trick on the LHS. Then you can use the triangle inequality (which will give you [itex] 1+|x+y| \leq 1 + |x| +|y|[/itex]) to compare [itex] -\frac{1}{1+|x+y|}[/itex] and [itex]-\frac{1}{1+|x|+|y|}[/itex]. Then you can use the first trick backwards, which should lead to the result. And hopefully I haven't made any mistakes with my inequalities!
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